[BZOJ] 1563: [NOI2009]诗人小G

1D/1D的方程，代价函数是一个p次函数，典型的决策单调性

用单调队列（其实算单调栈）维护决策点，优化转移

复杂度\(O(nlogn)\)

#include<iostream>
#include<cstring>
#include<cstdio>
 
using namespace std;
 
const int MAXN = 100005;
typedef long double ll;
 
struct Node{
    int l,r,p;
    Node(int _l=0,int _r=0,int _p=0){l=_l;r=_r;p=_p;}
}q[MAXN];
 
ll f[MAXN],s[MAXN];
int T,n,len;
 
ll p;
ll qpow(ll x,int y){
    if(x<0)x=-x;
    ll ret=1ll;
    while(y){
        if(y&1)ret*=x;
        x*=x;
        y>>=1;
    }
    return ret;
}
ll calc(int x,int y){return f[x]+qpow(s[y]-s[x]-len,p);}
 
int fnd(const Node &x,int y){
    int l=x.l,r=x.r+1,p=x.p,mid,ret=l;
    while(l<r){
        mid=(l+r)>>1;
        if(calc(y,mid)<=calc(p,mid))r=mid;
        else l=mid+1;
    }
    return l;
}
 
char str[32];
void solve(){
    cin>>n>>len>>p;len++;
    s[0]=0;f[0]=0;
    for(int i=1;i<=n;i++){
        scanf("%s",str);
        s[i]=s[i-1]+strlen(str)+1;
    }
    int h=1,t=0;
    q[++t]=Node(1,n,0);
    for(int i=1;i<=n;i++){
        while(h<=t&&q[h].r<i)h++;
        f[i]=calc(q[h].p,i);
        if(calc(i,n)<=calc(q[t].p,n)){
            while(h<=t&&calc(i,q[t].l)<calc(q[t].p,q[t].l))t--;
            if(h>t){q[++t]=Node(i+1,n,i);continue;}
            int x=fnd(q[t],i);
            q[t].r=x-1;q[++t]=Node(x,n,i);
        }
    }
    if(f[n]>1e18){puts("Too hard to arrange");return;}
    printf("%.0Lf\n",f[n]);
}
 
int main(){
    // freopen("out.out","w",stdout);
    cin>>T;
    while(T--)solve(),puts("--------------------");
}